The metal ions that have the calculated spin-only magnetic moment value of 4.9 B.M. are :

A. $\mathrm{Cr}^{2+}$

B. $\mathrm{Fe}^{2+}$

C. $\mathrm{Fe}^{3+}$

D. $\mathrm{Co}^{2+}$

E. $\mathrm{Mn}^{3+}$

Choose the correct answer from the options given below:

<p>To determine which metal ions have a calculated spin-only magnetic moment of 4.9 Bohr Magnetons (B.M.), we can use the formula:</p> <p>$ \text{Magnetic Moment (M.M)} = \sqrt{n(n+2)} \, \text{B.M.} $</p> <p>where $ n $ is the number of unpaired electrons. </p> <p>Given that the magnetic moment is 4.9 B.M., we can set up the equation:</p> <p>$ 4.9 = \sqrt{n(n+2)} $</p> <p>Solving this equation, we find that $ n = 4 $. </p> <p>Let's analyze each metal ion:</p> <p><p><strong>(A) ${}_{24} \mathrm{Cr}^{2+}$:</strong> Electronic configuration: $[\mathrm{Ar}] 3d^4$. This ion has 4 unpaired electrons. </p></p> <p><p><strong>(B) ${}_{26} \mathrm{Fe}^{2+}$:</strong> Electronic configuration: $[\mathrm{Ar}] 3d^6$. This ion also has 4 unpaired electrons. </p></p> <p><p><strong>(C) ${}_{26} \mathrm{Fe}^{3+}$:</strong> Electronic configuration: $[\mathrm{Ar}] 3d^5$. This ion has 5 unpaired electrons, which does not match the required $ n $ value of 4. </p></p> <p><p><strong>(D) ${}_{27} \mathrm{Co}^{2+}$:</strong> Electronic configuration: $[\mathrm{Ar}] 3d^7$. This ion has 3 unpaired electrons. </p></p> <p><p><strong>(E) ${}_{25} \mathrm{Mn}^{3+}$:</strong> Electronic configuration: $[\mathrm{Ar}] 3d^4$. This ion has 4 unpaired electrons.</p></p> <p>Based on the number of unpaired electrons, the metal ions with a spin-only magnetic moment of approximately 4.9 B.M. are $\mathrm{Cr}^{2+}$, $\mathrm{Fe}^{2+}$, and $\mathrm{Mn}^{3+}$.</p>